Finding the minimizer using fminunc.

If I have function:

fv = @(x) x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));

How would I use the function fminunc to find the minimizer rather than the minimum value?

Answer 1

[Torsten]

fv = @(x) x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));
[minimizer,minimum_value] = fminunc(fv,[1; 1])

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

minimizer = 2×1

0.8944 0.4472

minimum_value = 2.5279

[Howie]

So does this function just give you the minimizer as a 2x1?

[Torsten]

What else do you expect for a function with two variables to be optimized ?

An analytic expression for x1 and x2 ?

[Howie]

thank you!

Answer 2

[Walter Roberson]

If you need to find a maximum rather than a minimum then construct

nfv = @(x) -fv(x)

now minimize nfv using fminunc.

If you need to display what the maximum is remember to evaluate fv at the location of the maximum.

[Howie]

Sorry, I needed to find the minimizer rather than the minimum value Nor the maximum!